https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Mistymathmusic&feedformat=atomAoPS Wiki - User contributions [en]2024-03-29T14:37:42ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=Berkeley_mini_Math_Tournament&diff=204794Berkeley mini Math Tournament2023-11-20T08:29:49Z<p>Mistymathmusic: /* Structure */</p>
<hr />
<div>The '''Berkeley mini Math Tournament (BmMT)''' is a middle school math competition hosted by the Bay Area Math Tournament Group. It is a team competition, with each team having up to 5 participants.<br />
<br />
{{Contest Info|name=BmMT|region=California|type=Free Response|difficulty=1 - 3.5|breakdown=<u>Individual</u>: 2-2.5 (Problems 1-10), 3-3.5 (Problems 11-20)<br><u>Speed</u>: 1-2.5<br><u>Team</u>: 2-2.5 (Problems 1-10), 2.5-3.5 (Problems 11-20)}}<br />
<br />
==Structure==<br />
<br />
There are four rounds in BmMT: Puzzle, Individual, Team, and Relay.<br />
<br />
In the Puzzle Round, teams will work together to solve various puzzles, all of which follow a certain theme or style.<br />
<br />
In the Individual Round, each contestant will solve 20 problems alone in 60 minutes. <br />
<br />
In the Team Round, teams will work together to solve 20 problems in 60 minutes. <br />
<br />
In the Relay Round, teams will work together to solve 20 problems in 40 minutes, where some problems' answers depend on the answers to previous problems.<br />
<br />
All problems are free response, and legible writing is needed. All answers must be exact and simplified. Also, calculators are not allowed in the competition.<br />
<br />
==Content==<br />
<br />
The test mostly covers material from the Common Core State Standards, but with more critical thinking required. Additionally, problems cover quadratics, combinatorics, and divisibility. Trigonometry and logarithms will not be on the BmMT.<br />
<br />
==See Also==<br />
* [https://berkeley.mt/ Official BmMT Website]<br />
** [https://berkeley.mt/archives/ Past Problems]<br />
* [[Berkeley Math Tournament]], a competition for high schoolers also hosted by Berkeley<br />
<br />
[[Category:Mathematics competitions]]<br />
[[Category:Introductory mathematics competitions]]</div>Mistymathmusichttps://artofproblemsolving.com/wiki/index.php?title=Berkeley_mini_Math_Tournament&diff=204793Berkeley mini Math Tournament2023-11-20T08:28:37Z<p>Mistymathmusic: /* Structure */</p>
<hr />
<div>The '''Berkeley mini Math Tournament (BmMT)''' is a middle school math competition hosted by the Bay Area Math Tournament Group. It is a team competition, with each team having up to 5 participants.<br />
<br />
{{Contest Info|name=BmMT|region=California|type=Free Response|difficulty=1 - 3.5|breakdown=<u>Individual</u>: 2-2.5 (Problems 1-10), 3-3.5 (Problems 11-20)<br><u>Speed</u>: 1-2.5<br><u>Team</u>: 2-2.5 (Problems 1-10), 2.5-3.5 (Problems 11-20)}}<br />
<br />
==Structure==<br />
<br />
There are four rounds in BmMT: Puzzle, Individual, Team, and Relay.<br />
<br />
In the Puzzle Round, teams will work together to solve various puzzles, all of which follow a certain theme or style.<br />
In the Individual Round, each contestant will solve 20 problems alone in 60 minutes. <br />
In the Team Round, teams will work together to solve 20 problems in 60 minutes. <br />
In the Relay Round, teams will work together to solve 20 problems in 40 minutes, where some problems' answers depend on the answers to previous problems.<br />
<br />
All problems are free response, and legible writing is needed. All answers must be exact and simplified. Also, calculators are not allowed in the competition.<br />
<br />
==Content==<br />
<br />
The test mostly covers material from the Common Core State Standards, but with more critical thinking required. Additionally, problems cover quadratics, combinatorics, and divisibility. Trigonometry and logarithms will not be on the BmMT.<br />
<br />
==See Also==<br />
* [https://berkeley.mt/ Official BmMT Website]<br />
** [https://berkeley.mt/archives/ Past Problems]<br />
* [[Berkeley Math Tournament]], a competition for high schoolers also hosted by Berkeley<br />
<br />
[[Category:Mathematics competitions]]<br />
[[Category:Introductory mathematics competitions]]</div>Mistymathmusichttps://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12B_Problems/Problem_22&diff=1342222013 AMC 12B Problems/Problem 222020-09-28T18:27:03Z<p>Mistymathmusic: </p>
<hr />
<div>==Problem==<br />
Let <math>m>1</math> and <math>n>1</math> be integers. Suppose that the product of the solutions for <math>x</math> of the equation<br />
<cmath> 8(\log_n x)(\log_m x)-7\log_n x-6 \log_m x-2013 = 0 </cmath><br />
is the smallest possible integer. What is <math>m+n</math>?<br />
<br />
<math> \textbf{(A)}\ 12\qquad\textbf{(B)}\ 20\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 272 </math><br />
<br />
==Solution==<br />
Rearranging logs, the original equation becomes<br />
<cmath>\frac{8}{\log n \log m}(\log x)^2 - \left(\frac{7}{\log n}+\frac{6}{\log m}\right)\log x - 2013 = 0</cmath><br />
<br />
By Vieta's Theorem, the sum of the possible values of <math>\log x</math> is <math>\frac{\frac{7}{\log n}+\frac{6}{\log m}}{\frac{8}{\log n \log m}} = \frac{7\log m + 6 \log n}{8} = \log \sqrt[8]{m^7n^6}</math>. But the sum of the possible values of <math>\log x</math> is the logarithm of the product of the possible values of <math>x</math>. Thus the product of the possible values of <math>x</math> is equal to <math>\sqrt[8]{m^7n^6}</math>.<br />
<br />
It remains to minimize the integer value of <math>\sqrt[8]{m^7n^6}</math>. Since <math>m, n>1</math>, we can check that <math>m = 2^2</math> and <math>n = 2^3</math> work. Thus the answer is <math>4+8 = \boxed{\textbf{(A)}\ 12}</math>.<br />
<br />
==Video Solution==<br />
For those who prefer a video solution: https://www.youtube.com/watch?v=vX0y9lRv9OM&t=312s<br />
<br />
== See also ==<br />
{{AMC12 box|year=2013|ab=B|num-b=21|num-a=23}}<br />
{{MAA Notice}}</div>Mistymathmusichttps://artofproblemsolving.com/wiki/index.php?title=2014_AMC_12B_Problems/Problem_15&diff=1340032014 AMC 12B Problems/Problem 152020-09-24T18:24:48Z<p>Mistymathmusic: </p>
<hr />
<div>==Problem==<br />
<br />
When <math>p = \sum\limits_{k=1}^{6} k \ln{k}</math>, the number <math>e^p</math> is an integer. What is the largest power of 2 that is a factor of <math>e^p</math>?<br />
<br />
<math> \textbf{(A)}\ 2^{12}\qquad\textbf{(B)}\ 2^{14}\qquad\textbf{(C)}\ 2^{16}\qquad\textbf{(D)}\ 2^{18}\qquad\textbf{(E)}\ 2^{20} </math><br />
<br />
==Solution==<br />
Let's write out the sum. Our sum is equal to <br />
<cmath>1 \ln{1} + 2 \ln{2} + 3 \ln{3} + 4 \ln{4} + 5 \ln{5} + 6 \ln{6} = </cmath><br />
<cmath>\ln{1^1} + \ln{2^2} + \ln{3^3} + \ln {4^4} + \ln{5^5} + \ln {6^6} = </cmath><br />
<cmath>\ln{(1^1\times2^2\times3^3\times4^4\times5^5\times6^6)}</cmath><br />
Raising <math>e</math> to the power of this quantity eliminates the natural logarithm, which leaves us with <br />
<cmath>e^p = 1^1\times2^2\times3^3\times4^4\times5^5\times6^6</cmath><br />
This product has <math>2</math> powers of <math>2</math> in the <math>2^2</math> factor, <math>4*2=8</math> powers of <math>2</math> in the <math>4^4</math> factor, and <math>6</math> powers of <math>2</math> in the <math>6^6</math> factor.<br />
This adds up to <math>2+8+6=16</math> powers of two which divide into our quantity, so our answer is <math>\boxed{\textbf{(C)}\ 2^{16}}</math><br />
<br />
==Video Solution==<br />
<br />
For those wanting a video: https://www.youtube.com/watch?v=iq2X86GFVBo<br />
== See also ==<br />
{{AMC12 box|year=2014|ab=B|num-b=14|num-a=16}}<br />
{{MAA Notice}}</div>Mistymathmusichttps://artofproblemsolving.com/wiki/index.php?title=2020_AMC_12B_Problems/Problem_17&diff=1331112020 AMC 12B Problems/Problem 172020-09-04T23:46:41Z<p>Mistymathmusic: </p>
<hr />
<div>==Problem==<br />
How many polynomials of the form <math>x^5 + ax^4 + bx^3 + cx^2 + dx + 2020</math>, where <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> are real numbers, have the property that whenever <math>r</math> is a root, so is <math>\frac{-1+i\sqrt{3}}{2} \cdot r</math>? (Note that <math>i=\sqrt{-1}</math>)<br />
<br />
<math>\textbf{(A) } 0 \qquad \textbf{(B) }1 \qquad \textbf{(C) } 2 \qquad \textbf{(D) } 3 \qquad \textbf{(E) } 4</math><br />
<br />
==Solution 1==<br />
Let <math>P(x) = x^5+ax^4+bx^3+cx^2+dx+2020</math>. We first notice that <math>\frac{-1+i\sqrt{3}}{2} = e^{i\frac{2\pi}{3}}</math>. That is because of Euler's Formula : <math>e^{ix} = cos(x) + i \cdot sin(x)</math>. <math>\frac{-1+i\sqrt{3}}{2}</math> = <math>-\frac{1}{2} + i \cdot \frac {sqrt{3}}{2}</math> = <math>cos(120) + i \cdot sin(120)</math> = <math>e^{i \cdot 120(degrees)}</math> = <math>e^{i \frac{2}{3} \pi (radians)}</math>. In order <math>r</math> to be a root of <math>P</math>, <math>re^{i\frac{2\pi}{3}}</math> must also be a root of P, meaning that 3 of the roots of <math>P</math> must be <math>r</math>, <math>re^{i\frac{2\pi}{3}}</math>, <math>re^{i\frac{4\pi}{3}}</math>. However, since <math>P</math> is degree 5, there must be two additional roots. Let one of these roots be <math>w</math>, if <math>w</math> is a root, then <math>we^{i\frac{2\pi}{3}}</math> and <math>we^{i\frac{4\pi}{3}}</math> must also be roots. However, <math>P</math> is a fifth degree polynomial, and can therefore only have <math>5</math> roots. This implies that <math>w</math> is either <math>r</math>, <math>re^{i\frac{2\pi}{3}}</math>, or <math>re^{i\frac{4\pi}{3}}</math>. Thus we know that the polynomial <math>P</math> can be written in the form <math>(x-r)^m(x-re^{i\frac{2\pi}{3}})^n(x-re^{i\frac{4\pi}{3}})^p</math>. Moreover, by Vieta's, we know that there is only one possible value for the magnitude of <math>r</math> as <math>||r||^5 = 2020</math>, meaning that the amount of possible polynomials <math>P</math> is equivalent to the possible sets <math>(m,n,p)</math>. In order for the coefficients of the polynomial to all be real, <math>n = p</math> due to <math>re^{i\frac{2\pi}{3}}</math> and <math>re^{i\frac{4\pi}{3}}</math> being conjugates and since <math>m+n+p = 5</math>, (as the polynomial is 5th degree) we have two possible solutions for <math>(m, n, p)</math> which are <math>(1,2,2)</math> and <math>(3,1,1)</math> yielding two possible polynomials. The answer is thus <math>\boxed{\textbf{(C) } 2}</math>.<br />
<br />
~Murtagh<br />
<br />
==Solution 2==<br />
<br />
<br />
Let <math>x_1=r</math>, then <math>x_2=(-1+i\sqrt{3})/2 r</math>, x_3=〖((-1+i√3)/2)〗^2 r=(-1-i√3)/2 r, x_4=〖((-1+i√3)/2)〗^3 r=r which means x_4 will be back to x_1<br />
Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2, and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of the different root is wrong.<br />
The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br />
<br />
~Yelong_Li<br />
<br />
==Video Solution==<br />
<br />
Link: https://www.youtube.com/watch?v=8V5l5jeQjNg<br />
<br />
==See Also==<br />
<br />
{{AMC12 box|year=2020|ab=B|num-b=16|num-a=18}}<br />
{{MAA Notice}}</div>Mistymathmusichttps://artofproblemsolving.com/wiki/index.php?title=2003_AMC_12A_Problems/Problem_24&diff=1321352003 AMC 12A Problems/Problem 242020-08-19T03:36:35Z<p>Mistymathmusic: /* Video Solution */</p>
<hr />
<div>== Problem ==<br />
<br />
If <math>a\geq b > 1,</math> what is the largest possible value of <math>\log_{a}(a/b) + \log_{b}(b/a)?</math><br />
<br />
<math><br />
\mathrm{(A)}\ -2 \qquad<br />
\mathrm{(B)}\ 0 \qquad<br />
\mathrm{(C)}\ 2 \qquad<br />
\mathrm{(D)}\ 3 \qquad<br />
\mathrm{(E)}\ 4<br />
</math><br />
<br />
== Solution ==<br />
<br />
Using logarithmic rules, we see that<br />
<br />
<cmath>\log_{a}a-\log_{a}b+\log_{b}b-\log_{b}a = 2-(\log_{a}b+\log_{b}a)</cmath><br />
<cmath>=2-(\log_{a}b+\frac {1}{\log_{a}b})</cmath><br />
<br />
Since <math>a</math> and <math>b</math> are both positive, using [[AM-GM]] gives that the term in parentheses must be at least <math>2</math>, so the largest possible values is <math>2-2=0 \Rightarrow \boxed{\textbf{B}}.</math><br />
<br />
Note that the maximum occurs when <math>a=b</math>.<br />
<br />
==Video Solution==<br />
<br />
The Link: https://www.youtube.com/watch?v=InF2phZZi2A&t=1s<br />
<br />
-MistyMathMusic<br />
<br />
== See Also ==<br />
{{AMC12 box|year=2003|ab=A|num-b=23|num-a=25}}<br />
{{MAA Notice}}</div>Mistymathmusichttps://artofproblemsolving.com/wiki/index.php?title=2003_AMC_12A_Problems/Problem_24&diff=1321342003 AMC 12A Problems/Problem 242020-08-19T03:36:20Z<p>Mistymathmusic: /* Video Solution */</p>
<hr />
<div>== Problem ==<br />
<br />
If <math>a\geq b > 1,</math> what is the largest possible value of <math>\log_{a}(a/b) + \log_{b}(b/a)?</math><br />
<br />
<math><br />
\mathrm{(A)}\ -2 \qquad<br />
\mathrm{(B)}\ 0 \qquad<br />
\mathrm{(C)}\ 2 \qquad<br />
\mathrm{(D)}\ 3 \qquad<br />
\mathrm{(E)}\ 4<br />
</math><br />
<br />
== Solution ==<br />
<br />
Using logarithmic rules, we see that<br />
<br />
<cmath>\log_{a}a-\log_{a}b+\log_{b}b-\log_{b}a = 2-(\log_{a}b+\log_{b}a)</cmath><br />
<cmath>=2-(\log_{a}b+\frac {1}{\log_{a}b})</cmath><br />
<br />
Since <math>a</math> and <math>b</math> are both positive, using [[AM-GM]] gives that the term in parentheses must be at least <math>2</math>, so the largest possible values is <math>2-2=0 \Rightarrow \boxed{\textbf{B}}.</math><br />
<br />
Note that the maximum occurs when <math>a=b</math>.<br />
<br />
==Video Solution==<br />
<br />
The Link: https://www.youtube.com/watch?v=InF2phZZi2A&t=1s<br />
-MistyMathMusic<br />
<br />
== See Also ==<br />
{{AMC12 box|year=2003|ab=A|num-b=23|num-a=25}}<br />
{{MAA Notice}}</div>Mistymathmusichttps://artofproblemsolving.com/wiki/index.php?title=2003_AMC_12A_Problems/Problem_24&diff=1321322003 AMC 12A Problems/Problem 242020-08-19T03:36:02Z<p>Mistymathmusic: </p>
<hr />
<div>== Problem ==<br />
<br />
If <math>a\geq b > 1,</math> what is the largest possible value of <math>\log_{a}(a/b) + \log_{b}(b/a)?</math><br />
<br />
<math><br />
\mathrm{(A)}\ -2 \qquad<br />
\mathrm{(B)}\ 0 \qquad<br />
\mathrm{(C)}\ 2 \qquad<br />
\mathrm{(D)}\ 3 \qquad<br />
\mathrm{(E)}\ 4<br />
</math><br />
<br />
== Solution ==<br />
<br />
Using logarithmic rules, we see that<br />
<br />
<cmath>\log_{a}a-\log_{a}b+\log_{b}b-\log_{b}a = 2-(\log_{a}b+\log_{b}a)</cmath><br />
<cmath>=2-(\log_{a}b+\frac {1}{\log_{a}b})</cmath><br />
<br />
Since <math>a</math> and <math>b</math> are both positive, using [[AM-GM]] gives that the term in parentheses must be at least <math>2</math>, so the largest possible values is <math>2-2=0 \Rightarrow \boxed{\textbf{B}}.</math><br />
<br />
Note that the maximum occurs when <math>a=b</math>.<br />
<br />
==Video Solution==<br />
<br />
The Link: https://www.youtube.com/watch?v=InF2phZZi2A&t=1s<br />
-MistyMathMusic<br />
<br />
== See Also ==<br />
{{AMC12 box|year=2003|ab=A|num-b=23|num-a=25}}<br />
{{MAA Notice}}</div>Mistymathmusichttps://artofproblemsolving.com/wiki/index.php?title=2003_AMC_12A_Problems/Problem_24&diff=1321302003 AMC 12A Problems/Problem 242020-08-19T03:35:41Z<p>Mistymathmusic: </p>
<hr />
<div>== Problem ==<br />
<br />
If <math>a\geq b > 1,</math> what is the largest possible value of <math>\log_{a}(a/b) + \log_{b}(b/a)?</math><br />
<br />
<math><br />
\mathrm{(A)}\ -2 \qquad<br />
\mathrm{(B)}\ 0 \qquad<br />
\mathrm{(C)}\ 2 \qquad<br />
\mathrm{(D)}\ 3 \qquad<br />
\mathrm{(E)}\ 4<br />
</math><br />
<br />
== Solution ==<br />
<br />
Using logarithmic rules, we see that<br />
<br />
<cmath>\log_{a}a-\log_{a}b+\log_{b}b-\log_{b}a = 2-(\log_{a}b+\log_{b}a)</cmath><br />
<cmath>=2-(\log_{a}b+\frac {1}{\log_{a}b})</cmath><br />
<br />
Since <math>a</math> and <math>b</math> are both positive, using [[AM-GM]] gives that the term in parentheses must be at least <math>2</math>, so the largest possible values is <math>2-2=0 \Rightarrow \boxed{\textbf{B}}.</math><br />
<br />
Note that the maximum occurs when <math>a=b</math>.<br />
<br />
==Video Solution==<br />
<br />
The Link: https://www.youtube.com/watch?v=InF2phZZi2A&t=1s<br />
<br />
== See Also ==<br />
{{AMC12 box|year=2003|ab=A|num-b=23|num-a=25}}<br />
{{MAA Notice}}</div>Mistymathmusichttps://artofproblemsolving.com/wiki/index.php?title=1962_AHSME_Problems/Problem_40&diff=1318991962 AHSME Problems/Problem 402020-08-16T06:01:11Z<p>Mistymathmusic: </p>
<hr />
<div>==Problem==<br />
The limiting sum of the infinite series, <math>\frac{1}{10} + \frac{2}{10^2} + \frac{3}{10^3} + \dots</math> whose <math>n</math>th term is <math>\frac{n}{10^n}</math> is: <br />
<br />
<math> \textbf{(A)}\ \frac{1}9\qquad\textbf{(B)}\ \frac{10}{81}\qquad\textbf{(C)}\ \frac{1}8\qquad\textbf{(D)}\ \frac{17}{72}\qquad\textbf{(E)}\ \text{larger than any finite quantity} </math><br />
<br />
==Solution==<br />
The series can be written as the following:<br />
<br />
<math>\frac{1}{10} + \frac{1}{10^2} + \frac{1}{10^3} + ...</math><br />
<br />
<math>+ \frac{1}{10^2} + \frac{1}{10^3} + \frac{1}{10^4} + ...</math><br />
<br />
<math>+ \frac{1}{10^3} + \frac{1}{10^4} + \frac{1}{10^5} + ...</math><br />
<br />
and so on.<br />
<br />
by using the formula for infinite geometric series <math>(\frac{a}{1-r})</math>,<br />
<br />
We can get <math>\frac{\frac{1}{10}}{1-\frac{1}{10}}</math> <math>+</math> <math>\frac{\frac{1}{10^2}}{1-\frac{1}{10}}</math> <math>+</math> <math>\frac{\frac{1}{10^3}}{1-\frac{1}{10}}</math> <math>+</math> ...<br />
Since they all have common denominators, we get <math>\frac{(\frac{1}{10} + \frac{1}{10^2} + \frac{1}{10^3})}{\frac{9}{10}}</math>.<br />
Using the infinite series formula again, we get <math>\frac{\frac{\frac{1}{10}}{1-\frac{1}{10}}}{\frac{9}{10}}</math> <math>=</math> <math>\frac{\frac{\frac{1}{10}}{\frac{9}{10}}}{\frac{9}{10}}</math> <math>=</math> <math>\frac{\frac{1}{9}}{\frac{9}{10}}</math> <math>=</math> <math>\boxed{ (B) \frac{10}{81}}</math><br />
<br />
==Solution 2==<br />
So.. we have the sum to be <math>\frac{1}{10}+\frac{2}{100}+\frac{3}{1000}</math>...<br />
Notice that this can be written as <br />
<math>\frac{1}{10}+\frac{0.2}{10}+\frac{0.03}{10}+\frac{0.004}{10}</math>.<br />
Now, it is trivial that the new fraction we seek is <math>\frac{1.234567891011......}{10}</math><br />
<br />
Testing the answer choices, we see that <math>\boxed{B}</math> is the correct answer.<br />
<br />
==Video Solution==<br />
Problem starts at 2:20 : https://www.youtube.com/watch?v=3PDZtddYQoM&t=5s</div>Mistymathmusichttps://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12B_Problems/Problem_16&diff=1315842017 AMC 12B Problems/Problem 162020-08-12T03:18:25Z<p>Mistymathmusic: </p>
<hr />
<div>==Problem 16==<br />
The number <math>21!=51,090,942,171,709,440,000</math> has over <math>60,000</math> positive integer divisors. One of them is chosen at random. What is the probability that it is odd?<br />
<br />
<math>\textbf{(A)}\ \frac{1}{21} \qquad \textbf{(B)}\ \frac{1}{19} \qquad \textbf{(C)}\ \frac{1}{18} \qquad \textbf{(D)}\ \frac{1}{2} \qquad \textbf{(E)}\ \frac{11}{21}</math><br />
<br />
==Solution==<br />
If a factor of <math>21!</math> is odd, that means it contains no factors of <math>2</math>. We can find the number of factors of two in <math>21!</math> by counting the number multiples of <math>2</math>, <math>4</math>, <math>8</math>, and <math>16</math> that are less than or equal to <math>21</math> (Legendre's Formula). After some quick counting we find that this number is <math>10+5+2+1 = 18</math>. If the prime factorization of <math>21!</math> has <math>18</math> factors of <math>2</math>, there are <math>19</math> choices for each divisor for how many factors of <math>2</math> should be included (<math>0</math> to <math>18</math> inclusive). The probability that a randomly chosen factor is odd is the same as if the number of factors of <math>2</math> is <math>0</math> which is <math>\boxed{\textbf{(B)}\frac{1}{19}}</math>.<br />
<br />
Solution by: vedadehhc<br />
<br />
==Solution 2==<br />
We can write <math>21!</math> as its prime factorization:<br />
<cmath>21!=2^{18}\times3^9\times5^4\times7^3\times11\times13\times17\times19</cmath><br />
<br />
Each exponent of these prime numbers are one less than the number of factors at play here. This makes sense; <math>2^{18}</math> is going to have <math>19</math> factors: <math>2^0, 2^1, 2^2,...\text{ }2^{18}</math>, and the other exponents will behave identically. <br />
<br />
In other words, <math>21!</math> has <math>(18+1)(9+1)(4+1)(3+1)(1+1)(1+1)(1+1)(1+1)</math> factors. <br />
<br />
We are looking for the probability that a randomly chosen factor of <math>21!</math> will be odd--numbers that do not contain multiples of <math>2</math> as factors.<br />
<br />
From our earlier observation, the only factors of <math>21!</math> that are even are ones with at least one multiplier of <math>2</math>, so our probability of finding an odd factor becomes the following:<br />
<cmath>P(\text{odd})=\dfrac{\text{number of odd factors}}{\text{number of all factors}}=\dfrac{(9+1)(4+1)(3+1)(1+1)(1+1)(1+1)(1+1)}{(18+1)(9+1)(4+1)(3+1)(1+1)(1+1)(1+1)(1+1)}=\dfrac{1}{(18+1)}=\boxed{\dfrac{1}{19}}</cmath><br />
<br />
Solution submitted by [[User:TrueshotBarrage|David Kim]]<br />
<br />
==Video Solution==<br />
<br />
https://youtu.be/ZLHNTSIcGM8<br />
<br />
-MistyMathMusic<br />
<br />
==See Also==<br />
{{AMC12 box|year=2017|ab=B|num-b=15|num-a=17}}<br />
{{MAA Notice}}<br />
<br />
[[Category:Introductory Combinatorics Problems]]<br />
[[Category:Introductory Probability Problems]]</div>Mistymathmusichttps://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12B_Problems/Problem_16&diff=1315832017 AMC 12B Problems/Problem 162020-08-12T03:17:51Z<p>Mistymathmusic: </p>
<hr />
<div>==Problem 16==<br />
The number <math>21!=51,090,942,171,709,440,000</math> has over <math>60,000</math> positive integer divisors. One of them is chosen at random. What is the probability that it is odd?<br />
<br />
<math>\textbf{(A)}\ \frac{1}{21} \qquad \textbf{(B)}\ \frac{1}{19} \qquad \textbf{(C)}\ \frac{1}{18} \qquad \textbf{(D)}\ \frac{1}{2} \qquad \textbf{(E)}\ \frac{11}{21}</math><br />
<br />
==Solution==<br />
If a factor of <math>21!</math> is odd, that means it contains no factors of <math>2</math>. We can find the number of factors of two in <math>21!</math> by counting the number multiples of <math>2</math>, <math>4</math>, <math>8</math>, and <math>16</math> that are less than or equal to <math>21</math> (Legendre's Formula). After some quick counting we find that this number is <math>10+5+2+1 = 18</math>. If the prime factorization of <math>21!</math> has <math>18</math> factors of <math>2</math>, there are <math>19</math> choices for each divisor for how many factors of <math>2</math> should be included (<math>0</math> to <math>18</math> inclusive). The probability that a randomly chosen factor is odd is the same as if the number of factors of <math>2</math> is <math>0</math> which is <math>\boxed{\textbf{(B)}\frac{1}{19}}</math>.<br />
<br />
Solution by: vedadehhc<br />
<br />
==Solution 2==<br />
We can write <math>21!</math> as its prime factorization:<br />
<cmath>21!=2^{18}\times3^9\times5^4\times7^3\times11\times13\times17\times19</cmath><br />
<br />
Each exponent of these prime numbers are one less than the number of factors at play here. This makes sense; <math>2^{18}</math> is going to have <math>19</math> factors: <math>2^0, 2^1, 2^2,...\text{ }2^{18}</math>, and the other exponents will behave identically. <br />
<br />
In other words, <math>21!</math> has <math>(18+1)(9+1)(4+1)(3+1)(1+1)(1+1)(1+1)(1+1)</math> factors. <br />
<br />
We are looking for the probability that a randomly chosen factor of <math>21!</math> will be odd--numbers that do not contain multiples of <math>2</math> as factors.<br />
<br />
From our earlier observation, the only factors of <math>21!</math> that are even are ones with at least one multiplier of <math>2</math>, so our probability of finding an odd factor becomes the following:<br />
<cmath>P(\text{odd})=\dfrac{\text{number of odd factors}}{\text{number of all factors}}=\dfrac{(9+1)(4+1)(3+1)(1+1)(1+1)(1+1)(1+1)}{(18+1)(9+1)(4+1)(3+1)(1+1)(1+1)(1+1)(1+1)}=\dfrac{1}{(18+1)}=\boxed{\dfrac{1}{19}}</cmath><br />
<br />
Solution submitted by [[User:TrueshotBarrage|David Kim]]<br />
<br />
==Video Solution==<br />
<br />
https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_16<br />
<br />
-MistyMathMusic<br />
<br />
==See Also==<br />
{{AMC12 box|year=2017|ab=B|num-b=15|num-a=17}}<br />
{{MAA Notice}}<br />
<br />
[[Category:Introductory Combinatorics Problems]]<br />
[[Category:Introductory Probability Problems]]</div>Mistymathmusichttps://artofproblemsolving.com/wiki/index.php?title=2015_AMC_12A_Problems/Problem_14&diff=1310872015 AMC 12A Problems/Problem 142020-08-07T22:41:29Z<p>Mistymathmusic: </p>
<hr />
<div>==Problem==<br />
<br />
<!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>What is the value of <math>a</math> for which <math>\frac{1}{\text{log}_2a} + \frac{1}{\text{log}_3a} + \frac{1}{\text{log}_4a} = 1</math>?<!-- don't remove the following tag, for PoTW on the Wiki front page--></onlyinclude><br />
<br />
<math> \textbf{(A)}\ 9\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 24\qquad\textbf{(E)}\ 36</math><br />
<br />
==Solution==<br />
We use the change of base formula to show that<br />
<cmath>\log_a b = \dfrac{\log_b b}{\log_b a} = \dfrac{1}{\log_b a}.</cmath><br />
Thus, our equation becomes<br />
<cmath>\log_a 2 + \log_a 3 + \log_a 4 = 1,</cmath><br />
which becomes after combining:<br />
<cmath>\log_a 24 = 1.</cmath><br />
Hence <math>a = 24</math>, and the answer is <math>\textbf{(D)}.</math><br />
<br />
==Video Solution==<br />
Video starts at 1:04, uses the above solution: https://www.youtube.com/watch?v=OJyWHzjiu2A&t=44s<br />
<br />
== See Also ==<br />
{{AMC12 box|year=2015|ab=A|num-b=13|num-a=15}}</div>Mistymathmusichttps://artofproblemsolving.com/wiki/index.php?title=2020_AMC_12B_Problems/Problem_22&diff=1307442020 AMC 12B Problems/Problem 222020-08-06T05:40:10Z<p>Mistymathmusic: </p>
<hr />
<div>==Problem 22==<br />
<br />
What is the maximum value of <math>\frac{(2^t-3t)t}{4^t}</math> for real values of <math>t?</math><br />
<br />
<math>\textbf{(A)}\ \frac{1}{16} \qquad\textbf{(B)}\ \frac{1}{15} \qquad\textbf{(C)}\ \frac{1}{12} \qquad\textbf{(D)}\ \frac{1}{10} \qquad\textbf{(E)}\ \frac{1}{9}</math><br />
<br />
==Solution 1==<br />
We proceed by using AM-GM. We get <math>\frac{(2^t-3t) + 3t}{2}</math> <math>\ge \sqrt{(2^t-3t)(3t)}</math>. Thus, squaring gives us that <math>4^{t-1} \ge (2^t-3t)(3t)</math>. Rembering what we want to find(divide by <math>4^t</math>), we get the maximal values as <math>\frac{1}{12}</math>, and we are done.<br />
<br />
==Solution 2==<br />
<br />
Set <math> u = t2^{-t}</math>. Then the expression in the problem can be written as <cmath>R = - 3t^24^{-t} + t2^{-t}= - 3u^2 + u = - 3 (u - 1/6)^2 + \frac{1}{12} \le \frac{1}{12} .</cmath> It is easy to see that <math>u =\frac{1}{6}</math> is attained for some value of <math>t</math> between <math>t = 0</math> and <math>t = 1</math>, thus the maximal value of <math>R</math> is <math>\textbf{(C)}\ \frac{1}{12}</math>.<br />
<br />
==Solution 3 (Calculus Needed)==<br />
<br />
We want to maximize <math>f(t) = \frac{(2^t-3t)t}{4^t} = \frac{t\cdot 2^t-3t^2}{4^t}</math>. We can use the first derivative test. Use quotient rule to get the following:<br />
<cmath><br />
\frac{(2^t + t\cdot \ln{2} \cdot 2^t - 6t)4^t - (t\cdot 2^t - 3t^2)4^t \cdot 2\ln{2}}{(4^t)^2} = 0 \implies 2^t + t\cdot \ln{2} \cdot 2^t - 6t = (t\cdot 2^t - 3t^2) 2\ln{2} <br />
</cmath><br />
<cmath><br />
\implies 2^t + t\cdot \ln{2}\cdot 2^t - 6t = 2t\ln{2} \cdot 2^t - 6t^2 \ln{2}<br />
</cmath><br />
<cmath><br />
\implies 2^t(1-t\cdot \ln{2}) = 6t(1 - t\cdot \ln{2}) \implies 2^t = 6t<br />
</cmath>Therefore, we plug this back into the original equation to get <math>\boxed{\textbf{(C)} \frac{1}{12}}</math><br />
<br />
~awesome1st<br />
<br />
==Solution 4==<br />
<br />
First, substitute <math>2^t = x (\log_2{x} = t)</math> so that <br />
<cmath><br />
\frac{(2^t-3t)t}{4^t} = \frac{x\log_2{x}-3(\log_2{x})^2}{x^2}<br />
</cmath><br />
<br />
Notice that <br />
<cmath><br />
\frac{x\log_2{x}-3(\log_2{x})^2}{x^2} = \frac{\log_2{x}}{x}-3\Big(\frac{\log_2{x}}{x}\Big)^2.<br />
</cmath><br />
<br />
When seen as a function, <math>\frac{\log_2{x}}{x}-3\Big(\frac{\log_2{x}}{x}\Big)^2</math> is a synthesis function that has <math>\frac{\log_2{x}}{x}</math> as its inner function.<br />
<br />
If we substitute <math>\frac{\log_2{x}}{x} = p</math>, the given function becomes a quadratic function that has a maximum value of <math>\frac{1}{12}</math> when <math>p = \frac{1}{6}</math>.<br />
<br />
<br />
Now we need to check if <math>\frac{\log_2{x}}{x}</math> can have the value of <math>\frac{1}{6}</math> in the range of real numbers.<br />
<br />
In the range of (positive) real numbers, function <math>\frac{\log_2{x}}{x}</math> is a continuous function whose value gets infinitely smaller as <math>x</math> gets closer to 0 (as <math>log_2{x}</math> also diverges toward negative infinity in the same condition). When <math>x = 2</math>, <math>\frac{\log_2{x}}{x} = \frac{1}{2}</math>, which is larger than <math>\frac{1}{6}</math>.<br />
<br />
Therefore, we can assume that <math>\frac{\log_2{x}}{x}</math> equals to <math>\frac{1}{6}</math> when <math>x</math> is somewhere between 1 and 2 (at least), which means that the maximum value of <math>\frac{(2^t-3t)t}{4^t}</math> is <math>\boxed{\textbf{(C)}\ \frac{1}{12}}</math>.<br />
<br />
==Solution 5==<br />
Let the maximum value of the function be <math>m</math>. Then we have <cmath>\frac{(2^t-3t)t}{4^t} = m \implies m2^{2t} - t2^t + 3t^2 = 0.</cmath><br />
Solving for <math>2^{t}</math>, we see <cmath>2^{t} = \frac{t}{m} \pm \frac{\sqrt{t^2 - 12mt^2}}{m} = \frac{t}{m} \pm \frac{t\sqrt{1 - 12m}}{m}.</cmath> We see that <math>1 - 12m \geq 0 \implies m \leq \frac{1}{12}.</math> Therefore, the answer is <math>\boxed{\textbf{(C)}\ \frac{1}{12}}</math>.<br />
<br />
==Video Solution==<br />
Problem starts at 2:10 in this video: https://www.youtube.com/watch?v=5HRSzpdJaX0<br />
<br />
-MistyMathMusic<br />
<br />
==See Also==<br />
<br />
{{AMC12 box|year=2020|ab=B|num-b=21|num-a=23}}<br />
{{MAA Notice}}</div>Mistymathmusic